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The dissociation equilibrium of a gas AB, can be represented a

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The dissociation equilibrium of a gas AB, can be represented as The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K,, and total pressure p is

(2 K_{p} / p)

(2 K_{p} / p)^{1 / 3}

(2 K_{p} / p)^{1 / 2}

(K_{p} / p)

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The dissociation equilibrium of a gas AB, can be represented as The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K,, and total pressure p is

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Text solutionVerified

Initial moles At equil. 2 A B_{2} (g) 1 2 (1 - x) \Leftrightarrow 2 A B (g) + 0 2 x B_{2} (g) 0 x

where, x = degree of dissociation
Total moles at equilibrium

= 2 - 2 x + 2 x + x = (2 + x)

So,

p_{A B_{2}} = \frac{2 ( 1 - x ) p}{( 2 + x )}, p_{A B} = \frac{2 x p}{( 2 + x )}

p_{B_{2}} = \frac{x p}{( 2 + x )}

K_{p} = \frac{( p _{A B} ) ^{2} ( p _{B_{2}} )}{( p _{A B_{2}} ) ^{2}} = \frac{( \frac{2 x p}{2 + x} ) ^{2} [ ( \frac{x}{2 + x} ) p ]}{[ ( \frac{2 ( 1 - x )}{( 2 + x )} ) p ] ^{2}}

= \frac{4 x ^{3} p ^{3}}{( 2 + x ) ^{3}} \times \frac{( 2 + x ) ^{2}}{p ^{4} 4 ( 1 - x ) ^{2}} = \frac{x ^{3} p}{( 2 + x ) ( 1 - x ) ^{2}}

= \frac{x ^{3} p}{2} [∵ x <<< 1 and 2]

x = (\frac{2 K _{p}}{p})^{1 / 3}

so,

(1 - x) \approx 1

(2 + x) \approx 2

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Question Text	The dissociation equilibrium of a gas AB, can be represented as The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K,, and total pressure p is
Answer Type	Text solution:1
Upvotes	150

The dissociation equilibrium of a gas AB, can be represented as The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K,, and total pressure p is

Text solutionVerified

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